1. A neutron is fired at some Uranium-235. Barium-141 and Krypton-92 are produced:

${\displaystyle _{0}^{1}n+_{92}^{235}U\to _{56}^{141}Ba+_{36}^{92}Kr+N_{0}^{1}n}$

How many neutrons are produced (i.e. what is the value of N)?

Consider the mass numbers:

${\displaystyle 1+235=141+92+N}$

${\displaystyle N+233=236}$

${\displaystyle N=3}$

2. What proportion of the neutrons produced must be absorbed in order to make the reaction stable?

Two thirds, assuming that Uranium-235 always produces three neutrons when split, which it doesn't.

3. What would happen if too many neutrons were absorbed?

The reaction would stop.

4. Alternatively, Uranium-235 can split into Xenon-140, two neutrons and another element. What is this element? (You will need to use a periodic table.)

Consider the mass numbers:

${\displaystyle 1+235=140+N+2}$

${\displaystyle N+142=236}$

${\displaystyle N=94}$

Then consider the atomic numbers:

${\displaystyle 0+92=54+n+2(0)}$

${\displaystyle n=38}$

So, the element is strontium-94:

${\displaystyle _{0}^{1}n+_{92}^{235}U\to _{54}^{140}Xe+_{38}^{94}Sr+2_{0}^{1}n}$